Because the speed is half the escape speed and he formula is the escape speed formula but usualy the height is set to infinity so we get the formula from the beggining but in his case we are being asked for he height so we need it

Staff: Mentor

Because the speed is half the escape speed and he formula is the escape speed formula but usualy the height is set to infinity so we get the formula from the beggining but in his case we are being asked for he height so we need it

Ah. So your v/2 is the initial velocity, Vesc/2. Got it.

Since [itex] V_{esc} = \sqrt{\frac{2 \mu}{r_e}}[/itex] you can plug 1/2 of that into the total mechanical energy formula I gave above to find the constant energy [itex]\xi[/itex] at launch time. With that energy you can use the same formula again to find r when the velocity becomes zero. Note that the total mechanical energy formula is just a statement of the conservation of energy, in this case kinetic and potential energy.

Just to add: The gravitational parameter for some object μ is observable in and of itself if that object has sub-objects in orbit about it. In many cases, scientists know μ to a very high degree of precision (e.g., eleven decimal places for the Sun). In comparison, G is arguably the least known of all key constants in physics (only four or so decimal places). In fact, the primary evidence for many astronomical objects' mass is μ/G for that object.

Bottom line: It's not just a matter of convenience to lump GM into a single parameter μ. In many cases, using μ rather than GM is conceptually more precise (G and M are known to four places, μ to nine or more) and more correct (M is just μ/G).